CHEMISTRY SK025 TOPIC 3.0ELECTROCHEMISTRY 2025/20261TUTORIAL 3.03.1: Galvanic cell & Nernst equation1. (a) Definitioni. Measure of the half-cell to attract electrons. Its unit is volt.ii. Voltage associated with a reduction process occur at an electrode, andmeasured under 1 atm at 25°C and concentration of electrolyte is 1 M.(b) How to obtained SRP from SHE• Standard reduction potential is the voltage associated with a reductionprocess occur at an electrode, and measured under 1 atm pressure at 25°Cand concentration of electrolyte is 1 M.• Standard hydrogen electrode is consisting of hydrogen gas that is bubbledinto a 1.0 M hydrochloric acid solution at 25oC. Under standard condition, thehydrogen potential is zero.• If one half-cell is connected to SHE, the measured emf@ voltage of the cell isthe sum of electrical potentials between two electrodes.• If one of these electrode potential is zero, the other electrode potentialcould be obtained.(c) i. cell diagramii. overall cell reactionAnode : Al(s)→ Al3+ (aq) + 3eCathode: Pb2+(aq) + 2e→Pb(s)overall cell reaction: 2Al(s) + 3Pb2+(aq) → 2Al3+ (aq) + 3Pb(s)iii. Eocell( ) ( ) 0.13 1.66= +1.53 Vo Ecell = − − −Salt bridgeAl(anode) Pb(cathode)Al(NO3)3 (aq,1.0 M) Pb(NO3)2 (aq,1.0 M)Ve-Voltmeter
CHEMISTRY SK025 TOPIC 3.0: ELECTROCHEMISTRY 2025/20262iv. Electrode that increase in weightCathode (Pb) because Pb2+ ions in the solution undergo reduction toform Pb atoms that deposit on the cathode.2. (a) oxidizing agentI2(s) + 2e → 2I-(aq)Fe3+ (aq) + e → Fe2+ (aq)= +0.24 VIt is spontaneous reaction. Fe3+ can oxidise I-(b) Oxidizing agenti. Au3+ii. O2 in acidic medium(c) reducing agenti. Liii. Fe2+3. (a) function of salt bridgeTo complete the circuit so that electric current can flow and to maintain electrical neutrality. Example of salt bridge: KCl, Na2SO4, NH4NO3, NaNO3(b) half-cell equationsAnode : Ag(s) → Ag+(aq) + eCathode : M3+ (aq) + 3e → M (s)(c) balance equation(d) oxidizing agentM3+(e) Eocell= +1.8 V( ) ( ) 0.77 0.53 o Ecell = + − + 33 ( ) ( ) 3 ( ) ( ) Ag s M aq Ag aq M s + + + → + ( ) 3 ( ) 0.8 o ocell M ME E = − + +3oM ME +
CHEMISTRY SK025 TOPIC 3.0: ELECTROCHEMISTRY 2025/202634. (a) equationSn2+(aq) + Ni(s) → Sn(s) + Ni2+(aq)(b) cell notationNi(s) | Ni2+ (aq, 1M) || Sn2+ (aq, 1M) | Sn (s)(c) Eocell= +0.114 V(d) Ecell= + 0.173 V(e) Increase Ecell• increasing the concentration of reactant [Sn2+]• decreasing the concentration of product [Ni2+]5. (a) Overall cell reactionanode : Zn(s) → Zn2+(aq) + 2ē cathode : 2H+(aq) + 2ē → H2(g)overall : Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)cell notation :(b) SpontaneityEocell = Eocathode – Eoanode = 0.00 – (-0.76) = 0.76 V E0cell = + ve( the reaction is spontaneous)(c) Reducing agent: H+(aq)Oxidizing agent: Zn(s)(d) Ecell= + 0.61 V( ) ( ) 0.136 0.250 o Ecell = − − − ( ) 0.0592 0.01 0.114 log2 1.0Ecell = + − 22 Zn s Zn aq H aq H g Pt s ( ) ( ) ( ) ( ) ( ) + + ( ) ( ) ( )( )240.0592 0.01 10.76 log2 3 10Ecell− = + −
CHEMISTRY SK025 4(e) EcellpH = 1.2[ H+] = 0.0631 M= + 0.75 VThe value of Ecellis higher.6. (a) Overall cell reactionanode : cathode : Overall cell reaction: (b) pH= +0.34 V[H+] = 4.317 x 10-3 MpH = 2.367. From equation,Anode:Cathode: (a) Eocell From SRP table, = +0.02 V( ) ( ) ( )( )20.0592 0.01 10.76 log2 0.0631Ecell = + − 2 H g H aq e ( ) 2 ( ) 2 → + +2 Cu aq e Cu s ( ) 2 ( ) ++ → 22 Cu aq H g Cu s H aq ( ) ( ) ( ) 2 ( ) + + + → + ( ) ( ) 0.34 0.00 o Ecell = + − ( ) ( ) ( )20.0592 0.48 0.34 log2 1.0 1H + + = + − 2( ) ( ) 2 → + +Zn s Zn aq e 3( ) 3 ( ) + Cr aq e Cr s + → 2oZn /Zn E 0.76 V += − 3oCr /Cr E 0.74 V += −= −o o o E E E cell cathode anode 3 2 / /= −+ +o oCr Cr Zn ZnE E= − − − ( ) ( ) 0.74 V 0.76 VTOPIC 3.0: ELECTROCHEMISTRY 2025/2026
CHEMISTRY SK025 5(b) Ecell Given:[Zn2+] = 0.009 M, [Cr3+] = 0.10 M = +0.0608 VCell notation(c) Equilibrium constantAt equilibrium, Ecell = 0 V k = 106.428. Acidic Electrolyte❖ Hydrogen gas enter the fuel cell at anode, while oxygen gas at cathode.❖ At anode, hydrogen is oxidized to form H+ by donating 2e.❖ The H+ ion pass through the electrolyte to the cathode.32ocell cell 230.0592 ZnE E lgn Cr++ = − ( ) ( )( )320.0592 0.009 M0.02 V log6 0.10 M = − 2 3 0 009 0 1 + + Zn s Zn aq M Cr aq M Cr s ( ) ( , . ) ( , . ) ( )32ocell cell 230.0592 ZnE E lgn Cr++ = − ( ) 0.0592 0 0.02 log6= − kTOPIC 3.0: ELECTROCHEMISTRY 2025/2026
CHEMISTRY SK025 6❖ The electrons flow through an external circuit to generate an electric current.❖ At cathode, oxygen is reduced by receiving the 4e- from anode side and react with H+to produce water and heat as by-product.❖ Overall chemical reaction as below:Anode: 2H2(g) → 4H+(aq) + 4éCathode: O2 (g) + 4H+(aq) + 4é → 2H2O(l)Overall reaction: 2H2 (g) + O2 (g) → 2H2O (l)3.2: Electrolytic cell1. (a) ComparisonDifference Galvanic cell Electrolytic cellEnergy conversionEnergy release by spontaneous redox reaction is converted to electrical energyElectrical energy is used to drive nonspontaneous redox reactionElectrodes Anode is negative terminalCathode is positive terminalAnode is positive terminalCathode is negative terminalTransfer of electrons Negative to positive Positive to negative(b) Predict product at anode and cathodeSet Cathode AnodeIAnswer : Cu (s)Reason : Since Cu(s) electrode is an active electrode, Cu2+ will undergo reduction to form Cu(s).Answer : Cu2+ (aq)Reason : Since Cu(s) electrode is an active electrode, Cu(s) will undergo oxidation to produce Cu2+(aq)IIAnswer: Cu (s)Reason: Cu2+(aq) which have more positive value of E°red will undergo reduction to form Cu(s).Answer: O2 (g)Reason: SO42-cannot undergo oxidation because the oxidation number of S is maximum. Therefore, H2O will undergo oxidation and produce O2.2 Cu aq e Cu s ( ) 2 ( ) ++ → 2 Cu s Cu aq e ( ) ( ) 2 → + +2 Cu aq e Cu s ( ) 2 ( ) ++ →TOPIC 3.0: ELECTROCHEMISTRY 2025/2026
CHEMISTRY SK025 7IIIAnswer: Cu (s)Reason: Cu2+(aq) which have more positive value of E°red will undergo reduction to form Cu(s).Answer: Cu2+ (aq)Reason: Since Cu(s) electrode is an active electrode, Cu(s) will undergo oxidation to produce Cu2+(aq)IV Answer: Pb (s)Reason: For PbBr2 (l), Pb2+ are attracted to cathode and undergo reduction.Pb2+(l) + 2e Pb (l)Answer: Br2 (I)Reason: Br- are attracted to anode and undergo oxidation.2. (a) Faraday’s first lawThe amount of a substance produced or consumed during electrolysis is directly proportional to the amount of electricity that passes through the electrical circuit of the cell.(b) i. mass of silver1 mol Ag ≡ 1 mol e ≡ 1 F ≡ 96500 C96500 C ≡ 1 mol Ag1000 C ≡ 0.0104 mol Ag∴ Mass of Ag = 1.12 gii. time neededn Ag = 0.0185 mol1 mol Ag ≡ 96500 C0.0185 mol Ag ≡ 1785.25 C1785.25 C = (2)tt = 892.63 s3. Anode : Cathode :Overall: (a) Volumen Au = 0.04701 mol4 mol Au ≡ 3 mol O20.04701 mol Au ≡ 0.0353 mol O22 2 2 ( ) ( ) 4 ( ) 4 H O l O g H aq e → + + +2 Cu aq e Cu s ( ) 2 ( ) ++ → 2 Cu s Cu aq e ( ) ( ) 2 → + +22 ( ) ( ) 2 Br l Br l e − → + Ag aq e Ag s ( ) ( ) ++ → 2 2 2 ( ) ( ) 4 ( ) 4 H O l O g H aq e → + + +3 Au aq e Au s ( ) 3 ( ) ++ →32 2 6 ( ) 4 ( ) 3 ( ) 12 ( ) 4 ( ) H O l Au aq O g H aq Au s + + + → + +TOPIC 3.0: ELECTROCHEMISTRY 2025/2026
CHEMISTRY SK025 8= 0.872 L(b) current1 mol O2 ≡ 4 mol e ≡ 4F ≡ 4(96500 C)0.0353 mol O2 ≡ 13625.8 C13625.8 C = Q(7200 s)Q = 1.89 A4. timeAnode : Cathode : n H2 = = 0.382 mol1 mol H2 ≡ 2 mol e ≡ 2F ≡2(96500 C)0.382 mol H2 ≡ 73726 C73726 C = (12)tt = 6143.83 s5. VolumeAnode : Cathode :Overall : n Ag = 9.453 x 10-3 mol4 mol Ag ≡ 1 mol O29.453 x 10-3 mol Ag ≡2.363 x 10-3 mol O2= 0.06151 L( ) ( ) ( )( )0.0353 0.08206 296.150.983V =2 2 2 ( ) ( ) 4 ( ) 4 H O l O g H aq e → + + +2 2 2 ( ) 2 ( ) 2 ( ) H O l e H g OH aq −+ → + ( ) ( )( ) ( )0.924 10.000.08206 295.15 2 2 2 ( ) ( ) 4 ( ) 4 H O l O g H aq e → + + +Ag aq e Ag s ( ) ( ) ++ → 2 2 2 ( ) 4 ( ) ( ) 4 ( ) 4 ( ) H O l Ag aq O g H aq Ag s + + + → + + ( )( ) ( )( )32.363 10 0.08206 298.150.971V−=0.9398TOPIC 3.0: ELECTROCHEMISTRY 2025/2026
CHEMISTRY SK025 9MEKA 3.01. (a) cell notationi. From overall equation,Anode:Cathode:ii. Based on SRP table,(Cathode)(Anode)Anode :Cathode:cell notation(b) Eocelli. Based on cell notation,Anode:Cathode:From SRP table, = (-0.14 V) – (-0.40 V)= 0.26 V2( ) ( ) 2 → + +Sn s Sn aq e 2 Cu aq e Cu aq ( ) ( ) + + + → 2 2 Sn s Sn aq Cu aq Cu aq Pt s ( ) ( ) ( ) ( ) ( ) + + +2oCl /Cl E 1.36 V − = + 3 2oFe /Fe E 0.77 V + + = + 2 3 ( ) ( ) Fe aq Fe aq e + + → + 2( ) 2 2 ( ) − Cl g e Cl aq + → 2 32 Pt s Fe aq Fe aq Cl aq Cl g Pt s ( ) ( ), ( ) ( ) ( ) ( ) + + −2( ) ( ) 2 → + + Cd s Cd aq e 2( ) 2 ( ) +Sn aq e Sn s + → 2oCd /Cd E 0.40 V += − 2oSn /Sn E 0.14 V += −= −o o o E E E cell cathode anode 2 2 / /= −+ +o oSn Sn Cd Cd E ETOPIC 3.0: ELECTROCHEMISTRY 2025/2026
CHEMISTRY SK025 10ii. Based on equation,Anode :Cathode :From SRP table, = 1.05 V(c) SpontaneityBased on equation,Anode:Cathode:From SRP table, = -4.37 V∴ E0cell = - ve (the reaction is non-spontaneous)2. (a) Oxidizing agent: substance that undergo reduction reactionReducing agent: substance that undergo oxidation reaction.i. Cl2(g) will accept electron to form Cl−(aq): oxidizing agentii. will accept electron to from Mn2+(aq): oxidizing agentiii. Ba(s) will released electrons to form Ba2+(aq): reducing agentiv. Mg(s) will released electrons to from Mg2+(aq): reducing agent2( ) ( ) 2 → + + Ni s Ni aq e ( ) ( ) + Ag aq e Ag s + →oAg /Ag E 0.80 V + = + 2oNi /Ni E 0.25 V += −= −o o o E E E cell cathode anode2/ /= −+ +o oAg Ag Ni NiE E = + − − ( ) ( ) 0.80 0.25 3( ) ( ) 3 → + + Au s Au aq e 2( ) 2 ( ) + Ca aq e Ca s + → 3oAu /Au E 1.50 V + = + 2oCa /Ca E 2.87 V += −= −o o o E E E cell cathode anode 2 3 / /= −+ +o oCa Ca Au AuE E = − − + ( ) ( ) 2.87 V 1.50 V MnO (aq) 4−TOPIC 3.0: ELECTROCHEMISTRY 2025/2026
CHEMISTRY SK025 11(b) Ascending order of oxidizing agent: Based on SRP value, (-ve to +ve)Ag+(aq)< NO3−(aq) < Cr2O72−(aq)3. cell potentialFrom cell notation,Anode :Cathode :Overall equation: = +1.41 V = 1.45 VE cell value is positive, the reaction is spontaneous4. equilibrium constantFrom equation,Anode :Cathode :From SRP table, = (-0.14 V) – (-0.44 V)oAg /Ag E 0.80 V + = + 2 37oCrO /Cr E 1.33 V − + = +3oNO /NO E 0.96 V − = + ( )3( ) ( ) 3 2 → + + Al s Al aq e ( )2( ) 2 ( ) 3 + Ni aq e Ni s + → 2 3 2 ( ) 3 ( ) 2 ( ) 3 ( ) + + Al s Ni aq Al aq Ni s + → + = −o o o E E E cell cathode anode 2 3 / /= −+ +o oNi Ni Al Al E E= − − − ( ) ( ) 0.25 V 1.66 V23ocell cell 320.0592 AlE E lgn Ni++ = − ( ) ( )( )230.0592 0.01 M1.41 V log6 1.0 M = + − 2( ) ( ) 2 → + + Fe s Fe aq e 2( ) 2 ( ) +Sn aq e Sn s + → 2oFe /Fe E 0.44 V += − 2oSn /Sn E 0.14 V += −= −o o o E E E cell cathode anode 2 2 / /= −+ +o oSn Sn Fe FeE ETOPIC 3.0: ELECTROCHEMISTRY 2025/2026
CHEMISTRY SK025 12 = 0.30 VAt equilibrium, Ecell = 0 Vk = 1.37 x 10105. half-cell equations and product(a) Molten NaCl,The electrolyte contain Na+(l) and Cl−(l)Cl−(l) will be attracted to anode. While Na+(l) will be attracted to cathode.Anode : chlorine gas evolvedCathode : sodium liquid (b) The electrolyte only contains H2O, Na+(aq) and Cl-(aq).Anode (+ve terminal): Clion and water are attracted to the anode. H2O willbe selectively oxidized at anode because of water is less positive than Cl-. Anode: oxygen gas evolvedCathode (-ve terminal): Na+ion and water are attracted to the cathode. Cathode: hydrogen gas evolved6. Half-equation at the cathode: Mg2+(l) + 2e Mg(s) 2ocell cell 20.0592 FeE E lgn Sn++ = − ( ) 0.0592 0 0.30 V log2= − k22 ( ) ( ) 2 Cl l Cl g e − → + Na l e Na l ( ) ( ) ++ → o EreductionTOPIC 3.0: ELECTROCHEMISTRY 2025/2026
CHEMISTRY SK025 13From half cell equation at cathode: 1 mol of Mg = 2 mol of e KUMBE 3.01. (a) Draw diagramFrom SRP table,(cathode)(anode)Anode Cathode(b) salt bridgeKCl or NH4NO3 or Na2SO4(c) cell notationAnode:Cathode:oAg /Ag E 0.80 V + = + 2oCu /Cu E 0.34 V + = + 2( ) ( ) 2 → + + Cu s Cu aq e ( ( ) ( ) 2)+ Ag aq e Ag s + → 2+ + Cu(s) Cu (aq) Ag (aq) Ag(s)VeeeeCu(s) Ag(s)Salt bridgeCu2+(aq) Ag+(aq)VoltmeterTOPIC 3.0: ELECTROCHEMISTRY 2025/2026
CHEMISTRY SK025 14(c) Eocell = +0.46 V2. (a) i) Reducing agent: Fe2+ ionOxidizing agent: Ag+ionii) Concentration on [Ag+]From cell notation,Anode :Cathode : Overall Equation : From SRP table, Given: Ecell = 0.015 V, [Fe2+] = 0.125 M, [Fe3+] = 0.068 M= 0.03 V [Ag+] = 0.3035 M(b) pH= −o o o E E E cell cathode anode2/ /= −+ +o oAg Ag Cu CuE E = + − + ( ) ( ) 0.80 0.34 2 3 ( ) ( ) Fe aq Fe aq e + + → + ( ) ( ) + Ag aq e Ag s + → 2 3 Fe aq Ag aq Fe aq Ag s ( ) ( ) ( ) ( ) + + ++ → +oAg /Ag E 0.80 V + = + 3 2oFe /Fe E 0.77 V + + = += −o o o E E E cell cathode anode 3 2 / /= −+ + +o oFe Fe Ag AgE E = + − + ( ) ( ) 0.8 V 0.77 V3ocell cell 20.0592 FeE E lgn Fe Ag++ + = − ( ) ( )( )0.0592 0.068 M0.015 V 0.03 V log1 0.125 M + = − AgTOPIC 3.0: ELECTROCHEMISTRY 2025/2026
CHEMISTRY SK025 15From cell notation,Anode :Cathode : Overall Equation: Given: = +0.76 V, Ecell = 0.54 V, [Zn2+] = 1 M, [H+] = x M, = 1 atm[H+] = 1.922×10-4 MpH = - log [H+] = - log (1.922×10-4 M) = 3.723. Effect on Ecell(a) decrease(b) decrease(c) increase(d) remain unchanged4. (a) Cell diagram,(b) Equation and productThe electrolyte only contains H2O, Na+(aq) and .Both H2O, and will be attracted to anode, but H2O will be selectively oxidized at anode because of water is less positive than and 2( ) ( ) 2 → + +Zn s Zn aq e 22 ( ) 2 ( ) + H aq e H g + → 22 Zn s H aq Zn aq H g ( ) 2 ( ) ( ) ( ) + + + → + o Ecell ( ) H2P( ) 22o Hcell cell 20.0592 Zn PE E lgn H++ = − ( ) ( )( )20.0592 1.0 M 1.0 atm0.54 V 0.76 V log2 + = − Hdilute Na 2SO4(aq)carbon(s)anode cathode+ -24( ) −SO aq 24( ) −SO aq o Ereduction24( ) −SO aqTOPIC 3.0: ELECTROCHEMISTRY 2025/2026
CHEMISTRY SK025 16sulphur atom in cannot be oxidised because the sulphur atom in anion is already at the maximum oxidation state +6.Anode:Oxygen gas evolvedBoth H2O, and Na+(aq) will be attracted to cathode, but H2O will be selectively reduced at cathode because of water is higher than Na+. Cathode:Hydrogen gas evolved5. (a) massGiven:I = 1.73 At = 2.05 hours = 7380 sQ = It = (1.73 A)(7380 s) = 12767.4 C96500 C ≡ 1 mol e12767.4 C ≡ 0.132 mol e1 mol e ≡ 1 mol Ag0.132 mol e ≡ 0.132 mol Ag∴ mass of Ag = (0.132 mol)(107.9 g/mol) = 14.28 g(b) currentGiven:Mass of Ag = 0.212 g, nAg = 1.96×10−3 molt = 1435 s1 mol Ag ≡ 1 mol e 1.96×10−3 mol Ag ≡ 1.96×10−3 mol e24( ) −SO aq 2 2 2 ( ) ( ) 4 ( ) 4 H O l O g H aq e → + + +o Ereduction 2 2 2 ( ) 2 ( ) 2 ( ) H O l e H g OH aq −+ → + Ag aq e Ag s ( ) ( ) ++ → Ag aq e Ag s ( ) ( ) ++ →TOPIC 3.0: ELECTROCHEMISTRY 2025/2026
CHEMISTRY SK025 171 mol e ≡ 96500 C1.96×10−3 mol e ≡ 189.603 CQ = It(189.603 C) = (I)(1435 s)I = 0.132 A6. a)b) Anode ( +ve terminal)Au, H2O and Clion will be attracted to the anode.Due to active electrode (Au) used in the cell, Au will be oxidized to formAu3+ ions.Anode electrode becomes thinner @ mass of Au electrode decreasesc) Mass Au deposited at cathode Q = It= 2.0 ( 1.5 hr x 60 x 60)= 10800 CFrom balance equation at cathode3 mol of e = 1 mol of Au3 (96500) = 1 mol of Au 10800 mol e = 1 x 108003(96500)= 0.0373 mol AuMass Au deposited = 0.0373 x 197= 7.348 gTOPIC 3.0: ELECTROCHEMISTRY 2025/2026